What is the integral of 1/(x^2sqrt(9-x^2)) ?

The integral of 1/(x^2sqrt(9-x^2)) is -(sqrt(9-x^2))/(9x)+C

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integral of 1/(x^2sqrt(9-x^2))

\int \frac{1}{x ^{2} 9 - x ^{2}} d x

- \frac{9 - x ^{2}}{9 x} + C

Solution steps

\int \frac{1}{x ^{2} 9 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{9 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{9 cos ( u ) sin ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{9} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{9} (- cot (u))

Substitute back

u = arcsin (\frac{1}{3} x)

= \frac{1}{9} (- cot (arcsin (\frac{1}{3} x)))

Simplify

\frac{1}{9} (- cot (arcsin (\frac{1}{3} x))) : - \frac{9 - x ^{2}}{9 x}

= - \frac{9 - x ^{2}}{9 x}

Add a constant to the solution

= - \frac{9 - x ^{2}}{9 x} + C

\frac{- x ^{2} - 6}{( x ^{2} - 6 ) ^{2}}

\frac{5 x}{x + 4} (1.1)

y^{^{''}} + 5 y^{^{'}} = e^{- t}

y^{^{'}} + y cos (x) = 2 cos (x)

y = \frac{10}{x ^{2} - 2 x - 24}, x = - 2, x = 2

What is the integral of 1/(x^2sqrt(9-x^2)) ?
The integral of 1/(x^2sqrt(9-x^2)) is -(sqrt(9-x^2))/(9x)+C